Chem Help - Enthalpy of neutralisation (1 Viewer)

[DheerChoudhury]

So basically the answer says -53kJ/mol, which I guess I'm not disputing...
I'm just confused; I thought at the last step, you include the molar ratio of the water (2)?
Any help or consolidation would be appreciated

 

[CM_Tutor]

The equation is

Ba(OH)₂ + 2CH₃COOH ---> Ba(CH₃CO₂)₂ + 2H₂O​

The proper equation for finding is

​

In this case, the only substance we have the chemical amount of is ethanoic acid, and thus we have

​

 

[jazz519]

So basically the answer says -53kJ/mol, which I guess I'm not disputing...
I'm just confused; I thought at the last step, you include the molar ratio of the water (2)?
Any help or consolidation would be appreciated

View attachment 32865

I think it's because of the way they define the enthalpy of neutralisation in the HSC. It is stated as the amount of heat energy released from a neutralisation reaction when 1 mole of water is produced at 25 C and 100 kpa.

So the definition they use is deltaH neut = -q / n(H2O)

Based on that if you have q = 3.98145 kJ

2CH3COOH(aq) + Ba(OH)2(aq) --> Ba(CH3COO)2(aq) + 2H2O(l)

n(CH3COOH) = cv = (1.5)(50/1000) = 0.075 moles
n(H2O) = n(CH3COOH) = 0.075 moles

deltaH neut = -3.98145 / 0.075 = -53 kJ/mol


In terms of answering your molar ratio comment. Yes that was taken but it's not a molar ratio of 2. Because comparing to the substance you have the moles of which is CH3COOH, the ratios are exactly the same. There is a 2:2 ratio so the moles of CH3COOH is the same as moles of water made

 

[CM_Tutor]

I am not going to argue with @jazz519 about syllabus definitions.

However, the equation (with enthalpy change) is

Ba(OH)₂ + 2CH₃COOH ---> Ba(CH₃CO₂)₂ + 2H₂O
​

If they want to define it in terms of per mole of water, the equation must be:

Ba(OH)₂ + CH₃COOH ---> Ba(CH₃CO₂)₂ + H₂O ​

which is really bad chemical practice as fractions in equations should never produce fractional atoms, which they do in this case.

If they argued that neutralisation was H⁺ + OH^-, they would be incorrectly representing ethanoic acid as strong, when it is not.

 

[DheerChoudhury]

I think it's because of the way they define the enthalpy of neutralisation in the HSC. It is stated as the amount of heat energy released from a neutralisation reaction when 1 mole of water is produced at 25 C and 100 kpa.

So the definition they use is deltaH neut = -q / n(H2O)

Based on that if you have q = 3.98145 kJ

2CH3COOH(aq) + Ba(OH)2(aq) --> Ba(CH3COO)2(aq) + 2H2O(l)

n(CH3COOH) = cv = (1.5)(50/1000) = 0.075 moles
n(H2O) = n(CH3COOH) = 0.075 moles

deltaH neut = -3.98145 / 0.075 = -53 kJ/mol


In terms of answering your molar ratio comment. Yes that was taken but it's not a molar ratio of 2. Because comparing to the substance you have the moles of which is CH3COOH, the ratios are exactly the same. There is a 2:2 ratio so the moles of CH3COOH is the same as moles of water made

Ohhhh right, yeah I just realised the silly error. >_<

 

[jazz519]

I am not going to argue with @jazz519 about syllabus definitions.

However, the equation (with enthalpy change) is

Ba(OH)₂ + 2CH₃COOH ---> Ba(CH₃CO₂)₂ + 2H₂O
​

If they want to define it in terms of per mole of water, the equation must be:

Ba(OH)₂ + CH₃COOH ---> Ba(CH₃CO₂)₂ + H₂O ​

which is really bad chemical practice as fractions in equations should never produce fractional atoms, which they do in this case.

If they argued that neutralisation was H⁺ + OH^-, they would be incorrectly representing ethanoic acid as strong, when it is not.

Yeah this is correct but I guess for purpose of HSC they just have to calculate like that as that's what taught in textbooks like Pearson Chemistry

 

[CM_Tutor]

Yeah this is correct but I guess for purpose of HSC they just have to calculate like that as that's what taught in textbooks like Pearson Chemistry

The HSC has already proven that it will mark wrong answers that come from textbooks where the textbook is in error - it came up years back with the incorrect handling of eutrophication in the orange yr 12 conquering chemistry book, and it was noted in the examiner's report.

If textbooks have things wrong, it is up to them to change to match the syllabus, not the other way around... and the correct handling of delta H has been a mess for ages.